Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 434: 13


$$A = 72$$

Work Step by Step

Find the intersction points: $$12-x^{2}=x{2}-6$$ $$18=2x^{2}$$ $$9=x^{2}$$ $$x =3, or x=-3$$ Build the integral now that we know that the "top" function is $(12-x^{2})$ and the "bottom" is $(x^{2}-6)$: $$A=\displaystyle\int_{-3}^{3} (12 - x^{2})-(x^{2}-6)dx$$ $$A= \displaystyle\int_{-3}^{3} (-2x^{2}+18)dx$$ $$A= \left[ \frac{-2x^{3}}{3}+18x\right]^{3}_{-3}$$ Evaluated on on $[-3,3]$ Solve and simplify $$A = \Bigg[ \Big(-\frac{2\times3^{3}}{3}+18\times3\Big)-\Big(-\frac{2\times-3^{3}}{3}+18\Big)\Bigg]$$ $$A=(-18+54)-(18-54)$$ $$A = 72$$
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